What is the extreme f(x)= x/(x^2-x+1) ?

The extreme f(x)= x/(x^2-x+1) is Minimum(-1,-1/3),Maximum(1,1)

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extreme f(x)= x/(x^2-x+1)

extreme points

f (x) = \frac{x}{x ^{2} - x + 1}

Minimum (- 1, - \frac{1}{3}), Maximum (1, 1)

Solution steps

f^{'} (x) = \frac{- x ^{2} + 1}{( x ^{2} - x + 1 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = - 1

into

\frac{x}{x ^{2} - x + 1} : - \frac{1}{3}

Minimum (- 1, - \frac{1}{3})

Plug

x = 1

into

\frac{x}{x ^{2} - x + 1} : 1

Maximum (1, 1)

Minimum (- 1, - \frac{1}{3}), Maximum (1, 1)

f (x) = - x^{2} + 6 x - 5

f (x) = (19 - t)^{\frac{1}{2}}

x^{2} - 14

3 e^{x}

What is the extreme f(x)= x/(x^2-x+1) ?
The extreme f(x)= x/(x^2-x+1) is Minimum(-1,-1/3),Maximum(1,1)